The cross-sectional area of water pipe entering the basement is 4 x `10^(-4) m^(2)`. The pressure at this point is 3 x `10^(5) Nm^(-2)` and the speed of water is 2 `ms^(-1)`. This pipe tapers to a cross-sectional area of 2 x `10^(-4)m^(2)` when it reaches the second floor 8 m above. Calculate the speed and pressure at the 2nd floor

Explanation

Since `A_(1)nu_(1)=A_(2)nu_(2)`
`nu_(2)=(2xx4xx10^(-4))/(2xx10^(-4))` = 4m/s
Using Bernoulli’s Theorem
`P_(2)=P_(1)+(1)/(2)rho(upsilon_(1)^(2)-upsilon_(2)^(2))+rhog(h_(1)-h_(2))`
`:. nu_(2)gtnu_(1)`
`h_(2)gth_(1)`
`=3xx10^(5)+(1)/(2)(1000)[(2)^(2)-(4)^(2)]-1000xx9.8xx8`
`=2.16xx10^(5)N//m^(2)`